#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 45 Maths Textbook Solution.

Answer:$\frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{3}}}$

Hint:

\begin{aligned} &\frac{d}{d x}(\text { constant })=0 \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

Given:

$y=\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}$

Solution:

Put,

$x=\cos 2\theta$

So,

\begin{aligned} y &=\tan ^{-1}\left[\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right] \\ &=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}\right) \end{aligned}

Using

\begin{aligned} &1+\cos 2 \theta=2 \cos ^{2} \theta \\ &y=\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}\right) \\ &y=\tan ^{-1}\left(\frac{\sqrt{2}(\cos \theta-\sin \theta)}{\sqrt{2}(\cos \theta+\sin \theta)}\right) \end{aligned}

Dividing numerator and denominator by $\cos \theta$

\begin{aligned} &y=\tan ^{-1}\left(\frac{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}\right) \\ &y=\tan ^{-1}\left(\frac{\cos \theta-\sin \theta}{\frac{\cos \theta}{\cos \theta+\sin \theta}} \cos \theta\right) \end{aligned}

Using

\begin{aligned} &\tan \theta=\frac{\sin \theta}{\cos \theta} \\ &y=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \\ &y=\tan ^{-1}\left(\frac{\tan \frac{n}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\right) \\ &y=\tan ^{-1}\left(\tan \left(\frac{n}{4}-\theta\right)\right) \\ &\text { Using } \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \end{aligned}

$y=\frac{\pi-0}{4}-\theta$                                                                                                                                        $\left [ Using\: \tan ^{-1}\left ( \tan \theta \right ) =\theta \: if\: \theta \varepsilon \left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]\right ]$

$y=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\text { Using } x=\cos 2 \theta]$

Differentiating it with respect to x

$\frac{\mathrm{dy}}{\mathrm{d} x}=0-\frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\{\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\cos ^{-1} \mathrm{x}\right)=\frac{-1}{\sqrt{1-\mathrm{x}}^{2}}\right\}$

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}